Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A 3×33\times3 inverse computed by row reducing [A∣I][A\mid I]

Example

For A=(110011001),A=\begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix}, row reduction of [A∣I3][A\mid I_3] gives A−1=(1−1101−1001).A^{-1}=\begin{pmatrix}1&-1&1\\0&1&-1\\0&0&1\end{pmatrix}.

Facts & Assumptions

Given: The displayed real matrix AA.

[L1]

Reducing [A∣I][A\mid I] to [I∣B][I\mid B] gives B=A−1B=A^{-1} (Row reducing [A∣I][A\mid I] yields [I∣A−1][I\mid A^{-1}] exactly when AA is invertible).

[L2]

Matrix multiplication is associative and unital (Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication).

Verification

technique · direct
1.1

Starting from [A∣I3][A\mid I_3], replace row 11 by row 11 minus row 22, then replace row 00 by row 00 minus the new row 11. The left block becomes I3I_3 and the right block becomes the displayed matrix BB.

L1algebra
2.1

Direct multiplication gives AB=I3AB=I_3 and BA=I3BA=I_3: the only off-diagonal sums are −1+1-1+1, 1−11-1, and −1+1-1+1. Hence B=A−1B=A^{-1}.

step 1.1L2algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

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Sources